Question 314232: Alicia has $7000 invested, some at 7% and the remainder at 8%. After one year, the interest from the 7% investment exceeds the interest from the 8% investment by $70. How much is invested at each rate?
Thanks in advance for any assistance!
Answer by Earlsdon(6294)   (Show Source):
You can put this solution on YOUR website! Let x = the amount invested at 7% and ($7000-x) the amount invested at 8%.
Let y = the interest earned on the 8% investment so y+70 is the interest earned on the 7% investment.
After changing the percentages to their decimal equivalents, this can be expressed as a system of equations:
1) 0.07x = y+70 and
2) 0.08(7000-x) = y Subtract equation 2) from equation 1) to get:
0.07x-0.08(7000-x) = 70 Simplify and solve for x.
0.07x-560+0.08x = 70
0.15x -560 = 70 Add 560 to both sides.
0.15x = 630 Divide both sides by 0.15
x = $4200 and
$7000-x = $2800, so...
$4,200.00 was invested at 7% and $2,800.00 was invested at 8%
Check:
7% of $4,200.00 = $294.00
8% of $2,800.00 = $224.00
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